What are the divisors of 1999?

1, 1999

There is a total of 2 positive divisors.
The sum of these divisors is 2000.
The arithmetic mean is 1000.

2 odd divisors

1, 1999

How to compute the divisors of 1999?

A number N is said to be divisible by a number M (with M non-zero) if, when we divide N by M, the remainder of the division is zero.

$N mod M = 0$

Brute force algorithm

We could start by using a brute-force method which would involve dividing 1999 by each of the numbers from 1 to 1999 to determine which ones have a remainder equal to 0.

$Remainder = N - (M \times ⌊ \frac{N}{M} ⌋)$

(where $⌊ \frac{N}{M} ⌋$ is the integer part of the quotient)

1999 / 1 = 1999 (the remainder is 0, so 1 is a divisor of 1999)
1999 / 2 = 999.5 (the remainder is 1, so 2 is not a divisor of 1999)
1999 / 3 = 666.33333333333 (the remainder is 1, so 3 is not a divisor of 1999)
...
1999 / 1998 = 1.0005005005005 (the remainder is 1, so 1998 is not a divisor of 1999)
1999 / 1999 = 1 (the remainder is 0, so 1999 is a divisor of 1999)

Improved algorithm using square-root

However, there is another slightly better approach that reduces the number of iterations by testing only integers less than or equal to the square root of 1999 (i.e. 44.710177812216). Indeed, if a number N has a divisor D greater than its square root, then there is necessarily a smaller divisor d such that:

$D \times d = N$

(thus, if $\frac{N}{D} = d$ , then $\frac{N}{d} = D$ )

1999 / 1 = 1999 (the remainder is 0, so 1 and 1999 are divisors of 1999)
1999 / 2 = 999.5 (the remainder is 1, so 2 is not a divisor of 1999)
1999 / 3 = 666.33333333333 (the remainder is 1, so 3 is not a divisor of 1999)
...
1999 / 43 = 46.488372093023 (the remainder is 21, so 43 is not a divisor of 1999)
1999 / 44 = 45.431818181818 (the remainder is 19, so 44 is not a divisor of 1999)